wastes with fuel value heating value estimate for waste-to-energy design Given fraction carbon (C) = 0.4070; fraction hydrogen (H) = 0.0800; fraction oxygen (O) = 0.2980; fraction sulfur (S) = 0.0049, determine the higher heating value of waste (HHV) in MJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Wastes with Fuel Value.
• Everything except HHV is given, so isolate HHV symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Wastes with fuel value are characterized by a Dulong-type formula estimating higher heating value from elemental composition.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

2. Step 2 — Rearrange symbolically for HHV:

   $$HHV = 45.87C+143.34\left(H-\dfrac{O}{8}\right)+9.4S$$

3. Step 3 — List the givens: fraction carbon (C) = 0.4070, fraction hydrogen (H) = 0.0800, fraction oxygen (O) = 0.2980, fraction sulfur (S) = 0.0049.

4. Step 4 — Substitute the given values:

   $$HHV = 45.870.4070+143.34\left(0.0800-\dfrac{0.2980}{8}\right)+9.40.0049$$

5. Step 5 — Evaluate:

   $$HHV = 24.8429\ \text{MJ/kg}$$

6. Step 6 — Check: returning HHV = 24.8429 MJ/kg to

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 49.6859 — kept a factor of two that cancels in the correct rearrangement.
• 12.4215 — dropped that same factor in the other direction.
• 27.3272 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastes with Fuel Value (Dulong formula)

municipal solid wastes with fuel value characterized by elemental analysis Given fraction hydrogen (H) = 0.0380; fraction oxygen (O) = 0.2880; fraction sulfur (S) = 0.0191; higher heating value of waste (HHV) = 14.6800 MJ/kg, determine the fraction carbon (C).

Start with the thinking

• The governing relation printed in this handbook section is Wastes with Fuel Value.
• Everything except C is given, so isolate C symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Wastes with fuel value are characterized by a Dulong-type formula estimating higher heating value from elemental composition.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{HHV-143.34\left(H-\dfrac{O}{8}\right)-9.4S}{45.87}$$

3. Step 3 — List the givens: fraction hydrogen (H) = 0.0380, fraction oxygen (O) = 0.2880, fraction sulfur (S) = 0.0191, higher heating value of waste (HHV) = 14.6800 MJ/kg.

4. Step 4 — Substitute the given values:

   $$C = \dfrac{14.6800-143.34\left(0.0380-\dfrac{0.2880}{8}\right)-9.40.0191}{45.87}$$

5. Step 5 — Evaluate:

   $$C = 0.3099$$

6. Step 6 — Check: returning C = 0.3099 to

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.6197 — kept a factor of two that cancels in the correct rearrangement.
• 0.1549 — dropped that same factor in the other direction.
• 0.3409 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastes with Fuel Value (Dulong formula)

wastes with fuel value HHV computed from carbon, hydrogen, oxygen, sulfur fractions Given fraction carbon (C) = 0.2110; fraction hydrogen (H) = 0.0990; fraction oxygen (O) = 0.2920; fraction sulfur (S) = 0.0143, determine the higher heating value of waste (HHV) in MJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Wastes with Fuel Value.
• Everything except HHV is given, so isolate HHV symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Wastes with fuel value are characterized by a Dulong-type formula estimating higher heating value from elemental composition.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

2. Step 2 — Rearrange symbolically for HHV:

   $$HHV = 45.87C+143.34\left(H-\dfrac{O}{8}\right)+9.4S$$

3. Step 3 — List the givens: fraction carbon (C) = 0.2110, fraction hydrogen (H) = 0.0990, fraction oxygen (O) = 0.2920, fraction sulfur (S) = 0.0143.

4. Step 4 — Substitute the given values:

   $$HHV = 45.870.2110+143.34\left(0.0990-\dfrac{0.2920}{8}\right)+9.40.0143$$

5. Step 5 — Evaluate:

   $$HHV = 18.7717\ \text{MJ/kg}$$

6. Step 6 — Check: returning HHV = 18.7717 MJ/kg to

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 37.5435 — kept a factor of two that cancels in the correct rearrangement.
• 9.3859 — dropped that same factor in the other direction.
• 20.6489 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastes with Fuel Value (Dulong formula)

wastes with fuel value heating value estimate for waste-to-energy design Given fraction hydrogen (H) = 0.0450; fraction oxygen (O) = 0.1810; fraction sulfur (S) = 0.0053; higher heating value of waste (HHV) = 5.1300 MJ/kg, determine the fraction carbon (C).

Start with the thinking

• The governing relation printed in this handbook section is Wastes with Fuel Value.
• Everything except C is given, so isolate C symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Wastes with fuel value are characterized by a Dulong-type formula estimating higher heating value from elemental composition.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{HHV-143.34\left(H-\dfrac{O}{8}\right)-9.4S}{45.87}$$

3. Step 3 — List the givens: fraction hydrogen (H) = 0.0450, fraction oxygen (O) = 0.1810, fraction sulfur (S) = 0.0053, higher heating value of waste (HHV) = 5.1300 MJ/kg.

4. Step 4 — Substitute the given values:

   $$C = \dfrac{5.1300-143.34\left(0.0450-\dfrac{0.1810}{8}\right)-9.40.0053}{45.87}$$

5. Step 5 — Evaluate:

   $$C = 0.0408$$

6. Step 6 — Check: returning C = 0.0408 to

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0817 — kept a factor of two that cancels in the correct rearrangement.
• 0.0204 — dropped that same factor in the other direction.
• 0.0449 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastes with Fuel Value (Dulong formula)

municipal solid wastes with fuel value characterized by elemental analysis Given fraction carbon (C) = 0.2500; fraction hydrogen (H) = 0.0720; fraction oxygen (O) = 0.1690; fraction sulfur (S) = 0.0113, determine the higher heating value of waste (HHV) in MJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Wastes with Fuel Value.
• Everything except HHV is given, so isolate HHV symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Wastes with fuel value are characterized by a Dulong-type formula estimating higher heating value from elemental composition.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

2. Step 2 — Rearrange symbolically for HHV:

   $$HHV = 45.87C+143.34\left(H-\dfrac{O}{8}\right)+9.4S$$

3. Step 3 — List the givens: fraction carbon (C) = 0.2500, fraction hydrogen (H) = 0.0720, fraction oxygen (O) = 0.1690, fraction sulfur (S) = 0.0113.

4. Step 4 — Substitute the given values:

   $$HHV = 45.870.2500+143.34\left(0.0720-\dfrac{0.1690}{8}\right)+9.40.0113$$

5. Step 5 — Evaluate:

   $$HHV = 18.8661\ \text{MJ/kg}$$

6. Step 6 — Check: returning HHV = 18.8661 MJ/kg to

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 37.7323 — kept a factor of two that cancels in the correct rearrangement.
• 9.4331 — dropped that same factor in the other direction.
• 20.7528 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastes with Fuel Value (Dulong formula)

wastes with fuel value HHV computed from carbon, hydrogen, oxygen, sulfur fractions Given fraction hydrogen (H) = 0.0820; fraction oxygen (O) = 0.1820; fraction sulfur (S) = 0.0099; higher heating value of waste (HHV) = 13.0400 MJ/kg, determine the fraction carbon (C).

Start with the thinking

• The governing relation printed in this handbook section is Wastes with Fuel Value.
• Everything except C is given, so isolate C symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Wastes with fuel value are characterized by a Dulong-type formula estimating higher heating value from elemental composition.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{HHV-143.34\left(H-\dfrac{O}{8}\right)-9.4S}{45.87}$$

3. Step 3 — List the givens: fraction hydrogen (H) = 0.0820, fraction oxygen (O) = 0.1820, fraction sulfur (S) = 0.0099, higher heating value of waste (HHV) = 13.0400 MJ/kg.

4. Step 4 — Substitute the given values:

   $$C = \dfrac{13.0400-143.34\left(0.0820-\dfrac{0.1820}{8}\right)-9.40.0099}{45.87}$$

5. Step 5 — Evaluate:

   $$C = 0.0971$$

6. Step 6 — Check: returning C = 0.0971 to

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.1942 — kept a factor of two that cancels in the correct rearrangement.
• 0.0486 — dropped that same factor in the other direction.
• 0.1068 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastes with Fuel Value (Dulong formula)

wastes with fuel value heating value estimate for waste-to-energy design Given fraction carbon (C) = 0.4020; fraction hydrogen (H) = 0.0270; fraction oxygen (O) = 0.2260; fraction sulfur (S) = 0.0098, determine the higher heating value of waste (HHV) in MJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Wastes with Fuel Value.
• Everything except HHV is given, so isolate HHV symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Wastes with fuel value are characterized by a Dulong-type formula estimating higher heating value from elemental composition.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

2. Step 2 — Rearrange symbolically for HHV:

   $$HHV = 45.87C+143.34\left(H-\dfrac{O}{8}\right)+9.4S$$

3. Step 3 — List the givens: fraction carbon (C) = 0.4020, fraction hydrogen (H) = 0.0270, fraction oxygen (O) = 0.2260, fraction sulfur (S) = 0.0098.

4. Step 4 — Substitute the given values:

   $$HHV = 45.870.4020+143.34\left(0.0270-\dfrac{0.2260}{8}\right)+9.40.0098$$

5. Step 5 — Evaluate:

   $$HHV = 18.3527\ \text{MJ/kg}$$

6. Step 6 — Check: returning HHV = 18.3527 MJ/kg to

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 36.7054 — kept a factor of two that cancels in the correct rearrangement.
• 9.1763 — dropped that same factor in the other direction.
• 20.1880 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastes with Fuel Value (Dulong formula)

municipal solid wastes with fuel value characterized by elemental analysis Given fraction hydrogen (H) = 0.0720; fraction oxygen (O) = 0.3400; fraction sulfur (S) = 0.0045; higher heating value of waste (HHV) = 17.2000 MJ/kg, determine the fraction carbon (C).

Start with the thinking

• The governing relation printed in this handbook section is Wastes with Fuel Value.
• Everything except C is given, so isolate C symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Wastes with fuel value are characterized by a Dulong-type formula estimating higher heating value from elemental composition.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{HHV-143.34\left(H-\dfrac{O}{8}\right)-9.4S}{45.87}$$

3. Step 3 — List the givens: fraction hydrogen (H) = 0.0720, fraction oxygen (O) = 0.3400, fraction sulfur (S) = 0.0045, higher heating value of waste (HHV) = 17.2000 MJ/kg.

4. Step 4 — Substitute the given values:

   $$C = \dfrac{17.2000-143.34\left(0.0720-\dfrac{0.3400}{8}\right)-9.40.0045}{45.87}$$

5. Step 5 — Evaluate:

   $$C = 0.2819$$

6. Step 6 — Check: returning C = 0.2819 to

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.5637 — kept a factor of two that cancels in the correct rearrangement.
• 0.1409 — dropped that same factor in the other direction.
• 0.3101 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastes with Fuel Value (Dulong formula)

wastes with fuel value HHV computed from carbon, hydrogen, oxygen, sulfur fractions Given fraction carbon (C) = 0.4310; fraction hydrogen (H) = 0.0210; fraction oxygen (O) = 0.2840; fraction sulfur (S) = 0.0063, determine the higher heating value of waste (HHV) in MJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Wastes with Fuel Value.
• Everything except HHV is given, so isolate HHV symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Wastes with fuel value are characterized by a Dulong-type formula estimating higher heating value from elemental composition.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

2. Step 2 — Rearrange symbolically for HHV:

   $$HHV = 45.87C+143.34\left(H-\dfrac{O}{8}\right)+9.4S$$

3. Step 3 — List the givens: fraction carbon (C) = 0.4310, fraction hydrogen (H) = 0.0210, fraction oxygen (O) = 0.2840, fraction sulfur (S) = 0.0063.

4. Step 4 — Substitute the given values:

   $$HHV = 45.870.4310+143.34\left(0.0210-\dfrac{0.2840}{8}\right)+9.40.0063$$

5. Step 5 — Evaluate:

   $$HHV = 17.7508\ \text{MJ/kg}$$

6. Step 6 — Check: returning HHV = 17.7508 MJ/kg to

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 35.5015 — kept a factor of two that cancels in the correct rearrangement.
• 8.8754 — dropped that same factor in the other direction.
• 19.5258 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastes with Fuel Value (Dulong formula)

wastes with fuel value heating value estimate for waste-to-energy design Given fraction hydrogen (H) = 0.0930; fraction oxygen (O) = 0.3040; fraction sulfur (S) = 0.0043; higher heating value of waste (HHV) = 32.4200 MJ/kg, determine the fraction carbon (C).

Start with the thinking

• The governing relation printed in this handbook section is Wastes with Fuel Value.
• Everything except C is given, so isolate C symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Wastes with fuel value are characterized by a Dulong-type formula estimating higher heating value from elemental composition.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

2. Step 2 — Rearrange symbolically for C:

   $$C = \dfrac{HHV-143.34\left(H-\dfrac{O}{8}\right)-9.4S}{45.87}$$

3. Step 3 — List the givens: fraction hydrogen (H) = 0.0930, fraction oxygen (O) = 0.3040, fraction sulfur (S) = 0.0043, higher heating value of waste (HHV) = 32.4200 MJ/kg.

4. Step 4 — Substitute the given values:

   $$C = \dfrac{32.4200-143.34\left(0.0930-\dfrac{0.3040}{8}\right)-9.40.0043}{45.87}$$

5. Step 5 — Evaluate:

   $$C = 0.5340$$

6. Step 6 — Check: returning C = 0.5340 to

   $$HHV = 45.87 C + 143.34\left(H - \dfrac{O}{8}\right) + 9.4 S$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1.0681 — kept a factor of two that cancels in the correct rearrangement.
• 0.2670 — dropped that same factor in the other direction.
• 0.5874 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastes with Fuel Value (Dulong formula)